GIVEN A LIST OF REALS CAN YOU FIND A MISSING REAL ?

0.000
0.1?1
0.000

DIAG = 0.0?0
ADIAG = 0.1?1

CANTORS PROOF FAILS

ALL INFINITE STRINGS WITH AN UNKNOWN SYMBOL ? ARE LISTABLE

====================================================

LET POWERSET(N)
= {
{ 1 2 }
{ 1 3 5 }
...
}

REPRESENT { 1 3 5 } AS
1eS
3eS
5eS

NOW REPRESENT {1 3 5} AS
1eS
2!eS
3eS
4!eS
5eS
6!eS
....

CALL THE POWERSET(N)
{e !e}^N

OR SIMPLY
{T F}^N

THEN USE A TERNARY POWERSET
{T F ?}^N

EG.
SOME DEFINITIONS OF PRIME ALLOW 1 e PRIMES
PRIMES = {1? 2 3 5 7...}

1 ?e PRIMES
2 e PRIMES
3 e PRIMES
4 !e PRIMES
....

====================================================

let LIST(f(x)) be some list of some or all of the elements of f(x)

POWERSET PROOF
-----------------------------
ALL L
L = LIST( {T F}^N )
->
|L| < | {T F}^N | .........#

HOWEVER

EXIST L
L = LIST( {T F ?} ^N)
^ |L| = | {T F ?} ^ N |

The Triple Powerset can be listed

Since
| {T F}^N | < | {T F ?}^N |

The Powerset Proof can be safely ignored

# - this is not a cardinality inequality merely a statement that L is incomplete

On 2/16/23 10:39 PM, Graham Cooper wrote:
> GIVEN A LIST OF REALS CAN YOU FIND A MISSING REAL ?
>
> 0.000
> 0.1?1
> 0.000
>
> DIAG = 0.0?0
> ADIAG = 0.1?1
>
> CANTORS PROOF FAILS
>
> ALL INFINITE STRINGS WITH AN UNKNOWN SYMBOL ? ARE LISTABLE
>
>

Just because one disproof doesn't work doesn't mean that you have shown
you can do it.

If you include the ? in the shift 0->1->2->3->4->5->6->7->8->9->?->0

Then you can create the shifted number from the diagonal that does not
exist in the set.

Note, 0.1?1 is NOT the same as 0.101, you may say it "matches", but it
isn't the SAME NUMBER,

> ====================================================
>
>
> LET POWERSET(N)
> = {
> { 1 2 }
> { 1 3 5 }
> ..
> }

But a powerset HAS a definition, which isn't that.

You take a "Powerset" of a "Set", not "a number"

So, I am not sure what you are attempting to do.

You can't "define" things "by example", so you haven't actually defined
anything, so the rest of your arguement seems meaningles.

>
> REPRESENT { 1 3 5 } AS
> 1eS
> 3eS
> 5eS
>
> NOW REPRESENT {1 3 5} AS
> 1eS
> 2!eS
> 3eS
> 4!eS
> 5eS
> 6!eS
> ...
>
> CALL THE POWERSET(N)
> {e !e}^N
>
> OR SIMPLY
> {T F}^N
>
> THEN USE A TERNARY POWERSET
> {T F ?}^N
>
> EG.
> SOME DEFINITIONS OF PRIME ALLOW 1 e PRIMES
> PRIMES = {1? 2 3 5 7...}
>
> 1 ?e PRIMES
> 2 e PRIMES
> 3 e PRIMES
> 4 !e PRIMES
> ...
>
> ====================================================
>
>
> let LIST(f(x)) be some list of some or all of the elements of f(x)
>
>
> POWERSET PROOF
> -----------------------------
> ALL L
> L = LIST( {T F}^N )
> ->
> |L| < | {T F}^N | .........#
>
>
> HOWEVER
>
> EXIST L
> L = LIST( {T F ?} ^N)
> ^
> |L| = | {T F ?} ^ N |
>
> The Triple Powerset can be listed
>
> Since
> | {T F}^N | < | {T F ?}^N |
>
> The Powerset Proof can be safely ignored

Since your "powerset" isn't what is normally called the "powerset" I am
not sure what you are refering to.

>
>
> # - this is not a cardinality inequality merely a statement that L is incomplete

> > ====================================================
> >
> >
> > LET POWERSET(N)
> > = {
> > { 1 2 }
> > { 1 3 5 }
> > ..
> > }
> But a powerset HAS a definition, which isn't that.

Its just part of the TUTORIAL to introduce {TF?}^N as the TOTAL POWERSET

I could write

Let some (partial) enumeration of POWERSET(N) = { ...

OR

Consider a set of subsets

On 2/17/23 1:22 AM, Graham Cooper wrote:
>>> ====================================================
>>>
>>>
>>> LET POWERSET(N)
>>> = {
>>> { 1 2 }
>>> { 1 3 5 }
>>> ..
>>> }
>> But a powerset HAS a definition, which isn't that.
>
> Its just part of the TUTORIAL to introduce {TF?}^N as the TOTAL POWERSET
>
> I could write
>
> Let some (partial) enumeration of POWERSET(N) = { ...
>
> OR
>
> Consider a set of subsets
>

So, maybe you should try writing out actual definitions rather than just
ill defined examples.

Note, that since the 2^n subsets of an n element set become uncountable,
the 3^n "subsets" are even more uncountable.

A part of your issue is that "TOTAL POWER SUBSET" that has an optional
member is NOT the same subset as either of the "normal" subset that
contain or omit that member.

Math that tryies to claim that needs to loose some of the vital
properties of the natural numbers.

One big one is that if a == b, and b == c, then a == c.